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**Non-uniform Acceleration**

**Ques 21:** **x-coordinate of a particle moving along this axis is: x = (2 + t ^{2} + 2/3). Here, x is in metre and t in seconds. Find: (a) position of particle from where it started its journey, (b) initial velocity of particle and (c) acceleration of particle at t = 2s.**

(b) zero

(c) 26 ms

**Ques 22: The velocity of a particle moving in a straight line is decreasing at the rate of 3 m/s per metre of displacement at an instant when the velocity is 10 m/s. Determine the acceleration of the particle at this instant.****Ans: **- 30 ms^{-2}**Sol:** **Ques 23: The position o f a particle along a straight line is given by s = (t ^{3} - 9t^{2} - 15t) m, here t is in seconds. Determine its maximum acceleration during the time interval 0 < / < 10s.**

âˆ´

ie,

Acceleration (a) in the interval 0

a(at t = 10 s) = (6 x 10) - 18

= 42 m/s

(a) Find the initial speed v

(b) - 9.17 ms

or âˆ«dv = âˆ«(3 - 2t) dt

or v = 3t - t

or v = 3t - t

or âˆ«ds = âˆ« (3t - t

or

(a) Displacement at = Displacement at

â‡’ v

(b) v = 3 t - t

Velocity at t (= 5.00 s)

= 3.5 - 5

= 15 - 25 + 833

= - 9.167 m/s

Sol: v = 3t

âˆ´ âˆ«ds = âˆ«(3t

or s = t

or s = t

= 3.5 x 0.5

= 1.75 m/s

(b) Distance covered

= 4.036 m/s

i.e., va = 4 â€¦(i)

Now, at t = 2 s, v = 6 m/s

âˆ´

Thus, C = 10.

i.e., â€¦(ii)

âˆ´ at t = 3 s,

Substituting above found value of v in Eq.

(i),

i.e.,

= 0.603 m/s

âˆ´

or 3v + 2s = 60

or 3 v = 60 - 2s â€¦(i)

Differentiating above equation w.r.t. time t

or 3a = - 2v

or

or

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